We have received a very clearly explained solution to this, but unfortunately whoever sent it did not include their name. If it was you, let us know!

Let $ab=x$ and $cd=y$ (where $ab$ means $a$ as the tens digit, and $b$ as the ones digit, not $a$ times $b$).

\[\frac{abcd^2-cdab^2}{ab^2-cd^2} = \frac{(100x+y)^2 - (100y+x)^2}{x^2-y^2}\] \[= \frac{(10000x^2+200xy+y^2) - (10000y^2+200xy+x^2)}{x^2-y^2}\] \[= \frac{9999x^2 - 9999y^2}{x^2-y^2}\] \[= \frac{9999(x^2-y^2)}{x^2-y^2}\] \[ = 9999 \]

(since $x> y$ we are not dividing by zero)